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Question: The amplitude of an oscillating particle of mass 40 g decays to 10 cm from 100 cm in 20 s. Compute (...

The amplitude of an oscillating particle of mass 40 g decays to 10 cm from 100 cm in 20 s. Compute (a) Relaxation time and (b) Damping force when the velocity is 50 cm/s.

A

Relaxation time: τ=20ln⁡(10) s≈8.69 s\tau = \frac{20}{\ln(10)} \text{ s} \approx 8.69 \text{ s}; Damping force: Fd=200ln⁡(10) dynes≈461 dynesF_d = 200 \ln(10) \text{ dynes} \approx 461 \text{ dynes}

B

Relaxation time: τ=10ln⁡(20) s≈4.34 s\tau = \frac{10}{\ln(20)} \text{ s} \approx 4.34 \text{ s}; Damping force: Fd=100ln⁡(20) dynes≈300 dynesF_d = 100 \ln(20) \text{ dynes} \approx 300 \text{ dynes}

C

Relaxation time: τ=20ln⁡(10) s≈46.05 s\tau = 20 \ln(10) \text{ s} \approx 46.05 \text{ s}; Damping force: Fd=40ln⁡(10) dynes≈92.1 dynesF_d = 40 \ln(10) \text{ dynes} \approx 92.1 \text{ dynes}

D

Relaxation time: τ=100ln⁡(10) s≈43.43 s\tau = \frac{100}{\ln(10)} \text{ s} \approx 43.43 \text{ s}; Damping force: Fd=50ln⁡(10) dynes≈115.1 dynesF_d = 50 \ln(10) \text{ dynes} \approx 115.1 \text{ dynes}

Answer

(a) Relaxation time: τ=20ln⁡(10) s≈8.69 s\tau = \frac{20}{\ln(10)} \text{ s} \approx 8.69 \text{ s} (b) Damping force when velocity is 50 cm/s: $F_d = 200 \ln(10) \text{ dynes} \approx 461 \text{ dynes}

Explanation

Solution

The amplitude of a damped oscillating system decays exponentially with time, following the equation A(t)=A0e−γtA(t) = A_0 e^{-\gamma t}. The relaxation time τ\tau is the time taken for the amplitude to decay to 1/e1/e of its initial value, which means τ=1/γ\tau = 1/\gamma. The damping force is given by Fd=−bvF_d = -bv, where bb is the damping coefficient and vv is the velocity. The damping coefficient bb is related to the damping constant γ\gamma and the mass mm of the particle by b=2mγb = 2m\gamma.

  1. Calculate the damping constant (γ\gamma): Given: Initial amplitude A0=100A_0 = 100 cm, final amplitude A(t)=10A(t) = 10 cm, and time t=20t = 20 s. Using A(t)=A0e−γtA(t) = A_0 e^{-\gamma t}: 10 cm=100 cm×e−γ×20 s10 \text{ cm} = 100 \text{ cm} \times e^{-\gamma \times 20 \text{ s}} 0.1=e−20γ0.1 = e^{-20\gamma} Taking the natural logarithm of both sides: ln⁡(0.1)=−20γ\ln(0.1) = -20\gamma −ln⁡(10)=−20γ-\ln(10) = -20\gamma γ=ln⁡(10)20 s−1\gamma = \frac{\ln(10)}{20} \text{ s}^{-1}

  2. Calculate the relaxation time (τ\tau): The relaxation time is τ=1/γ\tau = 1/\gamma. τ=1ln⁡(10)20 s−1=20ln⁡(10) s\tau = \frac{1}{\frac{\ln(10)}{20} \text{ s}^{-1}} = \frac{20}{\ln(10)} \text{ s}

  3. Calculate the damping coefficient (bb): Given mass m=40m = 40 g. The relationship is b=2mγb = 2m\gamma. b=2×(40 g)×(ln⁡(10)20 s−1)b = 2 \times (40 \text{ g}) \times \left(\frac{\ln(10)}{20} \text{ s}^{-1}\right) b=4ln⁡(10) g/sb = 4 \ln(10) \text{ g/s}

  4. Calculate the damping force (FdF_d): The damping force is Fd=bvF_d = bv (magnitude). Given velocity v=50v = 50 cm/s. Fd=(4ln⁡(10) g/s)×(50 cm/s)F_d = (4 \ln(10) \text{ g/s}) \times (50 \text{ cm/s}) Fd=200ln⁡(10) g cm/s2F_d = 200 \ln(10) \text{ g cm/s}^2 Fd=200ln⁡(10) dynesF_d = 200 \ln(10) \text{ dynes}

Using ln⁡(10)≈2.3026\ln(10) \approx 2.3026: (a) τ≈202.3026≈8.686\tau \approx \frac{20}{2.3026} \approx 8.686 s (b) Fd≈200×2.3026≈460.52F_d \approx 200 \times 2.3026 \approx 460.52 dynes